Edexcel International A Level Maths · S1 · Statistics 1
The Normal distribution
28 free, high-quality past-paper-style questions on The Normal distribution for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 237 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-S1-6-B-01S1 · ASEasy6 marks
The random variable Z ~ N(0, 12).
(a)
Find P(Z < 1.36).
(1)
(b)
Find P(Z > −0.72).
(1)
(c)
Find P(−1.5 < Z < 0.85).
(2)
(d)
Find the value of a such that P(Z > a) = 0.025, and the value of b such that P(Z < b) = 0.2.
(2)
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Answer
Mark
Notes
Part (a)
Φ(1.36) = 0.9131
B1
Part (b)
by symmetry P(Z > −0.72) = Φ(0.72) = 0.7642
B1
Part (c)
Φ(0.85) − Φ(−1.5) = 0.8023 − (1 − 0.9332)
M1
= 0.7355 (awrt 0.736)
A1
Part (d)
a = 1.96
B1
b = −0.8416 (allow −0.84)
B1
Model sketch
Figure: N(0, 1) curve with P(−1.5 < Z < 0.85) shaded
Common error: Giving b = 0.8416. Only 20% of the distribution is below b, so b is below the mean of 0 and must be negative.
The height, H cm, of a tomato seedling three weeks after planting is modelled by H ~ N(18.0, 2.52).
(a)
Find the proportion of seedlings that are less than 15 cm tall.
(2)
(b)
Find P(H > 20).
(2)
(c)
The tallest 10% of seedlings are moved to larger pots. Find the least height of a seedling that is moved.
(2)
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Answer
Mark
Notes
Part (a)
P(Z < (15 − 18)/2.5) = P(Z < −1.2)
M1
1 − 0.8849 = 0.1151 = 0.115 (3 s.f.)
A1
Part (b)
P(Z > 0.8)
M1
1 − 0.7881 = 0.2119 = 0.212 (3 s.f.)
A1
Part (c)
(h − 18)/2.5 = 1.2816
M1
h = 18 + 1.2816 × 2.5 = 21.204 = 21.2 cm (3 s.f.)
A1
Model sketch
Figure: N(18.0, 2.5²) with the regions for parts (a) and (b) shaded
Exam tip: “The tallest 10%” is an upper-tail probability of 0.1, which is in the percentage points table: z = 1.2816 exactly, with no need to search the Φ table.
The mass, M grams, of a strawberry from one farm is modelled by M ~ N(μ, σ2). Figure 1 shows this model. The values marked on the axis are the mean and the points one, two and three standard deviations either side of it.
(a)
Use Figure 1 to write down the values of μ and σ.
(1)
(b)
Find the area of the shaded region, between 19 g and 31 g.
(3)
(c)
The heaviest 5% of strawberries are sold as “premium”. Find the least mass of a premium strawberry.
(2)
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Answer
Mark
Notes
Part (a)
μ = 24 and σ = 4 (both needed)
B1
Part (b)
standardises both: z = (19 − 24)/4 = −1.25 and (31 − 24)/4 = 1.75
M1
Φ(1.75) − (1 − Φ(1.25)) = 0.9599 − 0.1056
M1
= 0.8543 = 0.854 (3 s.f.)
A1
Part (c)
(m − 24)/4 = 1.6449
M1
m = 30.58 = 30.6 g (3 s.f.)
A1
Exam tip: The two ends of the shaded region are on opposite sides of the mean, so the region is Φ(upper) minus the lower tail. Sketch the curve and shade before you look anything up.
The reaction time, X milliseconds, of a driver to a warning light is modelled by X ~ N(μ, σ2). In Figure 1 the upper scale shows two reaction times, A and B, and the lower scale shows the standardised value z = (x − μ)/σ. Each of A and B lies directly above its standardised value.
(a)
Use the information in Figure 1 to find the values of μ and σ.
(3)
(b)
Find P(X < 300).
(2)
(c)
Write down the value of z that lies directly below a reaction time of 268 ms on the upper scale.
(1)
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Answer
Mark
Notes
Part (a)
(232 − μ)/σ = −1 and (292 − μ)/σ = 1.5
M1
subtracting: 60 = 2.5σ, so σ = 24
A1
μ = 232 + 24 = 256
A1
Part (b)
z = (300 − 256)/24 = 1.83 ft on their μ, σ
M1
Φ(1.83) = 0.9664 (awrt 0.966 or 0.967)
A1
Part (c)
(268 − 256)/24 = 0.5 ft
B1
Exam tip: Each matched pair (x, z) gives one equation x = μ + zσ. Two pairs give two simultaneous equations; subtract them to remove μ.
The box plot in Figure 1 summarises the commuting times, in minutes, of 120 cyclists. It is suggested that the commuting time, T minutes, of one of these cyclists can be modelled by T ~ N(31, 92).
(a)
Use Figure 1 to write down the interquartile range of the commuting times.
(1)
(b)
Using the model, find P(25 < T < 37).
(3)
(c)
Give two features of the box plot that support the use of the model.
(2)
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Answer
Mark
Notes
Part (a)
37 − 25 = 12 minutes
B1
Part (b)
z = ±6/9 = ±0.67
M1
2Φ(0.67) − 1 = 2 × 0.7486 − 1
M1
= 0.4972 (awrt 0.495 to 0.497)
A1
Part (c)
the median (31) is exactly halfway between the quartiles (25 and 37), so the middle of the data is symmetrical like a Normal distribution
B1
50% of the data lie between the quartiles, close to the model's value in part (b)
B1
Exam tip: Half of any data set lies between its quartiles. Comparing that 0.5 with the model's probability for the same interval is a quick test of a Normal model.
The time, T minutes, taken by a bus to complete its route is modelled by T ~ N(35, 62).
(a)
Find P(T > 44).
(2)
(b)
On one journey the bus has been travelling for 35 minutes and has not yet finished the route. Find the probability that the journey takes more than 41 minutes in total.
(3)
(c)
Three journeys are chosen at random. Find the probability that all three take less than 44 minutes.
(2)
(d)
State an assumption needed for your calculation in part (c).
Exam tip: In a conditional probability with Normal tails, check which event contains the other. Every journey over 41 minutes is also over 35 minutes, so the intersection is just P(T > 41).
A machine fills bags of rice. The mass, W grams, of rice in a bag is modelled by W ~ N(μ, 82). Exactly 2.5% of bags contain less than 500 g.
Figure: W ~ N(μ, 8²) with P(W < 500) shaded; not to scale
(a)
Find the value of μ.
(3)
(b)
The machine is adjusted so that the mean is 512 g and only 1% of bags contain less than 500 g. Find the new standard deviation.
(3)
(c)
After the adjustment, find the probability that a bag contains more than 520 g.
(2)
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Answer
Mark
Notes
Part (a)
z = −1.96
B1
(500 − μ)/8 = −1.96
M1
μ = 515.68 = 515.7 g
A1
Part (b)
z = −2.3263
B1
(500 − 512)/σ = −2.3263
M1
σ = 5.158… = 5.16 g (3 s.f.)
A1
Part (c)
P(Z > (520 − 512)/5.158) = P(Z > 1.55) ft on their σ
M1
1 − 0.9394 = 0.0606 (awrt 0.0605 or 0.0606)
A1
Common error: Using +1.96 in part (a). Only 2.5% of bags are below 500 g, so 500 is below the mean and the z-value is negative; the mean must come out above 500.
The time, X minutes, taken by adults to complete a logic puzzle is modelled by a Normal distribution with mean μ and standard deviation σ. 15% of adults take longer than 50 minutes and 10% take less than 30 minutes.
Figure: X ~ N(μ, σ²) with the two given tail areas shaded; not to scale
(a)
Write down two equations in μ and σ.
(2)
(b)
Find the values of μ and σ.
(4)
(c)
Find the probability that a randomly chosen adult takes less than 40 minutes.
(2)
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Answer
Mark
Notes
Part (a)
(50 − μ)/σ = 1.0364
B1
(30 − μ)/σ = −1.2816
B1
Part (b)
subtracts to eliminate μ: 20 = 2.318σ
M1
σ = 8.628… = 8.63 (3 s.f.)
A1
substitutes back, e.g. μ = 50 − 1.0364 × 8.628
M1
μ = 41.06… = 41.1 (3 s.f.)
A1
Part (c)
P(Z < (40 − 41.06)/8.628) = P(Z < −0.12) ft on their μ, σ
M1
1 − 0.5478 = 0.4522 (awrt 0.451 or 0.452)
A1
Exam tip: Both 0.15 and 0.1 are in the percentage points table, so use 1.0364 and 1.2816 to 4 d.p. Keep the unrounded σ in your calculator when you find μ.
At a health clinic the waiting times of patients have mean 14 minutes and standard deviation 10 minutes. A receptionist suggests modelling the waiting time, X minutes, by X ~ N(14, 102).
(a)
Using this model, find P(X < 0).
(2)
(b)
Use your answer to part (a) to comment on the suitability of the model.
(1)
(c)
At a second clinic the waiting time, Y minutes, is modelled by Y ~ N(25, 52). Find P(Y < 15).
(2)
(d)
Explain why the model for Y does not have the problem you described in part (b).
(2)
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Answer
Mark
Notes
Part (a)
P(Z < (0 − 14)/10) = P(Z < −1.4)
M1
1 − 0.9192 = 0.0808
A1
Part (b)
a waiting time cannot be negative, yet the model gives about 8% of patients a negative time, so the model is not suitable
B1
Part (c)
P(Z < −2)
M1
1 − 0.9772 = 0.0228
A1
Part (d)
0 is (0 − 25)/5 = −5, i.e. 5 standard deviations below the mean
M1
so P(Y < 0) is negligible (less than 0.0001) and the model gives essentially no negative times
A1
Model sketch
Figure: N(14, 10²) with P(X < 0) shaded
Exam tip: A Normal model for a quantity that cannot be negative is reasonable only when 0 is several standard deviations below the mean. Calculating P(X < 0) turns that judgement into a number.
The temperature, T °C, inside a refrigerator is recorded at random times. It is modelled by T ~ N(4.0, σ2). 5% of the recorded temperatures are above 6.0 °C.
Figure: T ~ N(4.0, σ²) with P(T > 6.0) shaded; not to scale
(a)
Find the value of σ.
(3)
(b)
Find the temperature below which 1% of the recorded temperatures lie.
(2)
(c)
Find P(3.0 < T < 5.0).
(3)
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Answer
Mark
Notes
Part (a)
z = 1.6449
B1
(6.0 − 4.0)/σ = 1.6449
M1
σ = 1.2158… = 1.22 °C (3 s.f.)
A1
Part (b)
(t − 4.0)/1.2158 = −2.3263 ft on their σ
M1
t = 1.171… = 1.17 °C (3 s.f.)
A1
Part (c)
z = ±1.0/1.2158 = ±0.82
M1
Φ(0.82) − Φ(−0.82) = 2 × 0.7939 − 1
M1
= 0.5878 (awrt 0.588 or 0.589)
A1
Exam tip: 3.0 and 5.0 are equally far from the mean, so P(3.0 < T < 5.0) = 2Φ(z) − 1: one look-up instead of two.
The distance, D metres, that a club athlete throws a javelin is modelled by a Normal distribution with mean μ and standard deviation σ. Figure 1 shows the graph of P(D ≤ d) against d for this model. Two points on the curve are marked.
(a)
Use the two marked points to write down two equations in μ and σ.
(2)
(b)
Find the values of μ and σ.
(4)
(c)
Find the probability that a throw is longer than 50 m.
(2)
(d)
Use Figure 1 to estimate the median throw, and explain why it should be close to your value of μ.
(1)
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Answer
Mark
Notes
Part (a)
(40 − μ)/σ = −1.2816
B1
(60 − μ)/σ = 1.6449
B1
Part (b)
subtracts to eliminate μ: 20 = 2.9265σ
M1
σ = 6.834… = 6.83 (3 s.f.)
A1
substitutes back, e.g. μ = 40 + 1.2816 × 6.834
M1
μ = 48.76… = 48.8 (3 s.f.)
A1
Part (c)
P(Z > (50 − 48.76)/6.834) = P(Z > 0.18) ft
M1
1 − 0.5714 = 0.4286 (awrt 0.428 or 0.429)
A1
Part (d)
reading across from 0.5: about 49 m (allow 48 to 50), which agrees with μ because a Normal distribution is symmetrical, so its median equals its mean
B1
Exam tip: A point (d, p) on a cumulative graph says P(D ≤ d) = p. For p = 0.10 and 0.95 the z-values come straight from the percentage points table; p below 0.5 gives a negative z.
The heights of two varieties of sunflower are modelled by Normal distributions. Variety A has height A cm, where A ~ N(150, 202), and variety B has height B cm, where B ~ N(μ, 122). Figure 1 shows the two distributions as curves P and Q. The proportion of plants taller than 180 cm is the same for both varieties.
(a)
Find P(A > 180).
(2)
(b)
Find the value of μ.
(3)
(c)
State which curve in Figure 1 represents variety B, giving a reason.
(1)
(d)
Find the probability that a plant of variety B is shorter than 150 cm, and compare it with the corresponding probability for variety A.
(2)
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Answer
Mark
Notes
Part (a)
P(Z > (180 − 150)/20) = P(Z > 1.5)
M1
1 − 0.9332 = 0.0668
A1
Part (b)
P(B > 180) = 0.0668 gives z = 1.5
B1
(180 − μ)/12 = 1.5
M1
μ = 162
A1
Part (c)
Q, because B has the smaller standard deviation (12 < 20), so its curve is narrower and taller (its mean, 162, is also further right)
B1
Part (d)
P(B < 150) = P(Z < −1) = 1 − 0.8413 = 0.1587 ft on their μ
B1
P(A < 150) = 0.5, so a variety A plant is far more likely to be under 150 cm
B1
Exam tip: Equal tail areas mean equal z-values: 180 is 1.5 standard deviations above the mean for both varieties, so 180 = μ + 1.5 × 12.
A school bus leaves its depot at the time shown in Figure 1. The time, T minutes, that it takes to travel from the depot to a stop is modelled by T ~ N(21, 2.52). The bus does not wait at the stop. Figure 1 also shows when two students, Priya and Tom, reach the stop on one morning. A student catches the bus if they reach the stop before it does.
(a)
Find the probability that Priya catches the bus.
(2)
(b)
Find the probability that Tom catches the bus.
(2)
(c)
The bus company changes the departure time so that the bus reaches the stop before 07:55 on only 1% of mornings. Keeping the same model for T, find the earliest departure time, to the nearest minute, that achieves this.
(3)
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Answer
Mark
Notes
Part (a)
Priya arrives 18 minutes after departure, so P(T > 18) = P(Z > −1.2)
M1
Φ(1.2) = 0.8849
A1
Part (b)
Tom arrives 25 minutes after departure: P(T > 25) = P(Z > 1.6)
M1
1 − 0.9452 = 0.0548
A1
Part (c)
need P(T < k) = 0.01 where k is the time from departure to 07:55; z = −2.3263
B1
k = 21 − 2.3263 × 2.5 = 15.18 minutes
M1
departure no earlier than 07:55 − 15.18 = 07:39.8, so the earliest whole minute is 07:40
A1
Exam tip: Turn clock times into minutes after departure before standardising: 07:48 is 18 minutes after 07:30.
The time, X seconds, that a customer spends on the checkout page of an online shop is modelled by a Normal distribution with mean μ and standard deviation σ. Figure 1 shows the proportions of customers in three time intervals.
(a)
Write down the percentage of customers who spend more than 60 seconds on the page.
(1)
(b)
Find the values of μ and σ.
(5)
(c)
Find the probability that a customer spends less than 45 seconds on the page.
(2)
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Answer
Mark
Notes
Part (a)
100 − 5 − 65 = 30%
B1
Part (b)
(36 − μ)/σ = −1.6449
B1
(60 − μ)/σ = 0.5244
B1
subtracts: 24 = 2.1693σ
M1
σ = 11.06… = 11.1 (3 s.f.)
A1
μ = 36 + 1.6449 × 11.06 = 54.19… = 54.2 (3 s.f.)
A1
Part (c)
P(Z < (45 − 54.20)/11.06) = P(Z < −0.83) ft
M1
1 − 0.7967 = 0.2033 (awrt 0.203)
A1
Common error: Using the 65% in the middle as a tail. The table needs areas in a tail: below 36 is 5% and above 60 is 30%, so 60 has z = 0.5244.
Figure 1 shows a column of a parcel locker, with a small compartment above a large one. Every parcel fits the width and depth of both compartments. The height, H cm, of a parcel is modelled by H ~ N(21, 62). A parcel goes in the small compartment if it fits; otherwise it goes in the large compartment if it fits; otherwise it is returned to the sender.
(a)
Use Figure 1 to find the internal height of the large compartment.
(1)
(b)
Find the probability that a parcel is returned to the sender.
(2)
(c)
The designer moves the shelf so that 80% of parcels fit in the small compartment. Find the least internal height of the small compartment, to the nearest millimetre.
(3)
(d)
With the shelf in its new position, find the probability that a parcel is returned to the sender.
(2)
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Answer
Mark
Notes
Part (a)
66 − 24 − 2 = 40 cm
B1
Part (b)
P(H > 40) = P(Z > 3.17) ft on their 40
M1
1 − 0.9992 = 0.0008
A1
Part (c)
z = 0.8416
B1
(h − 21)/6 = 0.8416, h = 26.05
M1
26.1 cm (rounded up so that at least 80% fit)
A1
Part (d)
large compartment height 66 − 2 − 26.1 = 37.9 cm, P(Z > (37.9 − 21)/6) = P(Z > 2.82) ft
M1
1 − 0.9976 = 0.0024
A1
Exam tip: When a length must be large enough for at least a given proportion, round up: 26.0 cm would hold slightly less than 80% of parcels.
A factory makes metal rods on two machines. The length of a rod made on machine A is A mm, where A ~ N(25.0, 0.42). The length of a rod made on machine B is B mm, where B ~ N(25.1, 0.252). A rod is accepted if its length is between 24.5 mm and 25.5 mm; otherwise it is rejected.
Figure: the length distributions for machines A and B, with the acceptance limits 24.5 mm and 25.5 mm
(a)
Find the probability that a rod made on machine A is accepted.
(3)
(b)
Find the probability that a rod made on machine B is accepted.
(3)
(c)
40% of the rods are made on machine A and the rest on machine B. Find the probability that a rod chosen at random is accepted.
(2)
(d)
A rod chosen at random is rejected. Find the probability that it was made on machine A.
(3)
(e)
The factory manager wants to reduce the number of rejected rods by adjusting one machine. State, with a reason, which machine should be adjusted.
(1)
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Answer
Mark
Notes
Part (a)
z = (25.5 − 25.0)/0.4 = 1.25 and (24.5 − 25.0)/0.4 = −1.25
M1
2Φ(1.25) − 1 = 2 × 0.8944 − 1
M1
= 0.7888 = 0.789 (3 s.f.)
A1
Part (b)
z = (25.5 − 25.1)/0.25 = 1.6 and (24.5 − 25.1)/0.25 = −2.4
M1
Φ(1.6) − (1 − Φ(2.4)) = 0.9452 − 0.0082
M1
= 0.9370 = 0.937 (3 s.f.)
A1
Part (c)
0.4 × 0.7888 + 0.6 × 0.9370 ft on their (a), (b)
M1
= 0.8777 = 0.878 (3 s.f.)
A1
Part (d)
P(A ∩ rejected) = 0.4 × (1 − 0.7888) = 0.08448
B1
divides by P(rejected) = 1 − 0.87772 = 0.12228
M1
0.08448/0.12228 = 0.6909 = 0.691 (3 s.f.)
A1
Part (e)
machine A: although it makes fewer rods, about 69% of rejected rods come from it, because its standard deviation is larger
B1
Exam tip: Draw a tree diagram: first branch machine A or B, second branch accepted or rejected, with the Normal probabilities on the second branches. Part (d) is then a standard conditional probability from the tree.
The length, L cm, of a fish of one species in a lake is modelled by a Normal distribution with mean μ and standard deviation σ. 20% of these fish are shorter than 24 cm and 5% are longer than 40 cm.
Figure: L ~ N(μ, σ²) with the two given tail areas shaded; not to scale
(a)
Find the values of μ and σ.
(5)
(b)
Anglers must return to the lake any fish shorter than 26 cm. Find the probability that a fish caught at random must be returned.
(2)
(c)
A fish that is caught is long enough to keep. Find the probability that it is longer than 40 cm.
(3)
(d)
Two fish are caught independently. Find the probability that exactly one of them must be returned.
(2)
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Answer
Mark
Notes
Part (a)
(24 − μ)/σ = −0.8416
B1
(40 − μ)/σ = 1.6449
B1
subtracts: 16 = 2.4865σ
M1
σ = 6.434… = 6.43 (3 s.f.)
A1
μ = 24 + 0.8416 × 6.434 = 29.41… = 29.4 (3 s.f.)
A1
Part (b)
P(Z < (26 − 29.41)/6.434) = P(Z < −0.53) ft
M1
1 − 0.7019 = 0.2981 (awrt 0.298)
A1
Part (c)
P(L > 40 | L > 26) = P(L > 40)/P(L > 26)
M1
= 0.05/(1 − 0.2981) ft
M1
= 0.0712 (awrt 0.0712)
A1
Part (d)
2 × 0.2981 × 0.7019 ft
M1
= 0.418 (3 s.f.)
A1
Common error: Using 1 − 0.05 = 0.95 as the numerator in part (c). The numerator is P(L > 40), which the stem gives directly as 0.05; only the denominator, P(L > 26) = 1 − 0.2981, needs a complement.
A takeaway restaurant records the time, D minutes, between an order being placed and the food being delivered. D is modelled by a Normal distribution with mean μ and standard deviation σ. 6.68% of deliveries take less than 34 minutes and 2.28% take more than 76 minutes.
Figure: D ~ N(μ, σ²) with the two given tail areas shaded; not to scale
(a)
Show that σ = 12 and find the value of μ.
(5)
(b)
Find P(D < 46).
(2)
(c)
Find the delivery time that is exceeded by 30% of deliveries.
(2)
(d)
The restaurant gives a refund on any delivery that takes more than 70 minutes. One evening it makes 500 deliveries. Estimate the number of refunds it gives.
(3)
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Answer
Mark
Notes
Part (a)
P(D < 34) = 0.0668 and Φ(1.5) = 0.9332, so z = −1.5 for 34
B1
P(D > 76) = 0.0228 and Φ(2) = 0.9772, so z = 2 for 76
B1
μ − 1.5σ = 34 and μ + 2σ = 76
M1
3.5σ = 42, so σ = 12
A1
μ = 52
A1
Part (b)
P(Z < −0.5)
M1
1 − 0.6915 = 0.3085 = 0.309 (3 s.f.)
A1
Part (c)
(d − 52)/12 = 0.5244
M1
d = 58.29… = 58.3 minutes (3 s.f.)
A1
Part (d)
P(D > 70) = P(Z > 1.5)
M1
= 0.0668
A1
500 × 0.0668 = 33.4, so about 33 refunds ft
A1
Exam tip: In a “show that” for σ, both z-values must appear with their signs and both equations must be written down. The printed answer only confirms your working.
Applicants for a job take an aptitude test. Their scores, S points, are modelled by S ~ N(58, 122).
(a)
Find P(40 < S < 70).
(3)
(b)
Applicants in the top 10% are invited straight to a final interview. Find the lowest score that earns this invitation.
(3)
(c)
Three applicants are chosen at random. Find the probability that exactly one of them scores more than 70 points.
(3)
(d)
The employer wants 75% of applicants to pass the test. Find the pass mark that achieves this, giving your answer to the nearest whole number.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
z = −1.5 and z = 1
M1
Φ(1.00) − (1 − Φ(1.5)) = 0.8413 − 0.0668
M1
= 0.7745 = 0.775 (3 s.f.)
A1
Part (b)
z = 1.2816
B1
(s − 58)/12 = 1.2816
M1
s = 73.37… = 73.4 points (3 s.f.)
A1
Part (c)
P(S > 70) = 1 − Φ(1.00) = 0.1587
B1
3 × 0.1587 × 0.84132, with the factor 3 for the three orders
M1
= 0.337 (3 s.f.)
A1
Part (d)
25% below the pass mark, so Φ(z) = 0.75 read in reverse: z = −0.67 (allow −0.674 or −0.6745)
B1
(p − 58)/12 = −0.67
M1
p = 49.96 (awrt 49.9 or 50.0), so 50 points
A1
Model sketch
Figure: N(58, 12²) with P(40 < S < 70) shaded
Exam tip: 0.25 is not in the percentage points table, so read Φ(z) = 0.75 from the main table: 0.7486 at z = 0.67 is the nearest entry. No interpolation is needed, and a calculator value of 0.6745 is equally acceptable.
this is lower than £1.15, so resetting is not worthwhile: the extra scrapped pins cost more than the regrinding saved ft
B1
Exam tip: Moving the mean away from one limit moves it towards the other. Always recompute both tail probabilities after a change before deciding whether it helps.
In a fairground game, darts are thrown at the board shown in Figure 1. Only the horizontal position of a dart matters, and the score is shown in each strip. For one player, the horizontal distance, X cm, of a dart to the right of the centre line (negative to the left) is modelled by X ~ N(0, σ2). 60% of this player's darts land in the central strip, which scores 10.
(a)
Show that σ = 4.75 to 3 significant figures.
(3)
(b)
Find the probability that a dart lands in a strip scoring 5.
(3)
(c)
Find the expected score for one dart.
(3)
(d)
A second player has the same value of σ but aims 2 cm to the right of the centre line, so that X ~ N(2, σ2). Find the probability that this player's dart lands in the central strip scoring 10.
z = (4 − 2)/4.753 = 0.42 and (−4 − 2)/4.753 = −1.26
M1
Φ(0.42) − (1 − Φ(1.26)) = 0.6628 − 0.1038
M1
= 0.5590 (awrt 0.559 or 0.560)
A1
Exam tip: With mean 0 the strips come in matching pairs, so work out one side and double it. Once the mean moves to 2 the symmetry is lost and each boundary needs its own z-value.
The amount, C mg, of active ingredient in a tablet is modelled by C ~ N(μ, σ2). A production batch is approved only if both of the requirements shown in Figure 1 are met.
(a)
For one batch μ = 250 and σ = 4. Show that this batch is approved.
(3)
(b)
Keeping σ = 4, find the range of values of μ for which a batch is approved.
(4)
(c)
Find the greatest value of σ, to 3 significant figures, for which a batch can be approved, and the value of μ that must then be used.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
P(C < 240) = P(Z < −2.5) = 1 − 0.9938 = 0.0062
M1
0.0062 ≤ 0.01, so the lower requirement is met
A1
P(C > 262) = P(Z > 3) = 1 − 0.9987 = 0.0013 ≤ 0.005, so both requirements are met and the batch is approved
the two limits must be at least 2.3263σ + 2.5758σ apart: 4.9021σ ≤ 262 − 240 = 22
M1
σ ≤ 4.4879…
A1
σ = 4.48 (rounded down, since 4.49 would break a requirement)
A1
μ = 240 + 2.3263 × 4.4879 = 250.44 (accept 250.42 to 250.46 with σ = 4.48; not 250.4, which breaks the lower requirement)
A1
Common error: Rounding the limits of μ the wrong way. 249.3 is below 249.3052, so a batch with that mean would fail; round a lower bound up and an upper bound down.
Figure 1 shows the mean daily maximum temperature, in °C, for each month in a coastal town. For each month from June to August, the daily maximum temperature T °C on a randomly chosen day is modelled by a Normal distribution with the mean shown in Figure 1 and standard deviation 2.5 °C.
(a)
Find the probability that on a randomly chosen day in July the maximum temperature is above 30 °C.
(3)
(b)
A “hot month” is one in which P(T > 30) is at least 0.1. Find the least mean temperature for a hot month, and hence state which of June, July and August are hot months.
(4)
(c)
In September the daily maximum temperature is also modelled by a Normal distribution with the mean shown in Figure 1, but with a different standard deviation σ. On 5% of September days the maximum temperature is below 18 °C. Find σ.
(3)
(d)
Find the probability that on a randomly chosen day in September the maximum temperature is above 28 °C.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
July mean = 27
B1
P(T > 30) = P(Z > (30 − 27)/2.5) = P(Z > 1.2)
M1
1 − 0.8849 = 0.1151
A1
Part (b)
z = 1.2816
B1
(30 − μ)/2.5 ≤ 1.2816
M1
μ ≥ 26.796, so the least mean is 26.8 °C
A1
July (27) and August (28) are hot months; June (25) is not ft
A1
Part (c)
September mean = 24 and z = −1.6449
B1
(18 − 24)/σ = −1.6449
M1
σ = 3.6476… = 3.65 °C (3 s.f.)
A1
Part (d)
z = (28 − 24)/3.6476 = 1.10 ft on their σ
M1
1 − 0.8643 = 0.1357 (awrt 0.136)
A1
Exam tip: “At least 0.1 above 30” means 30 must be no more than 1.2816 standard deviations above the mean. Writing the inequality before solving shows which way the answer goes.
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